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Difficulty: MediumStandard Enthalpy Changes and Hess's Law

The standard enthalpies of combustion of carbon, hydrogen, and propane are given below:

C(s)+O2(g)→CO2(g)ΔH∘=−393.5 kJ mol−1\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -393.5\text{ kJ mol}^{-1}
H2(g)+12O2(g)→H2O(l)ΔH∘=−285.8 kJ mol−1\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -285.8\text{ kJ mol}^{-1}
C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)ΔH∘=−2220.0 kJ mol−1\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \quad \Delta H^\circ = -2220.0\text{ kJ mol}^{-1}

What is the standard enthalpy of formation of propane, C3H8(g)\text{C}_3\text{H}_8(g)?

  1. −103.7 kJ mol−1-103.7\text{ kJ mol}^{-1}Answer
  2. B
    +103.7 kJ mol−1+103.7\text{ kJ mol}^{-1}
  3. C
    −4543.7 kJ mol−1-4543.7\text{ kJ mol}^{-1}
  4. D
    +4543.7 kJ mol−1+4543.7\text{ kJ mol}^{-1}

Answer

−103.7 kJ mol−1-103.7\text{ kJ mol}^{-1}
To calculate the standard enthalpy of formation of propane, combine the enthalpy changes for burning 3 moles of carbon and 4 moles of hydrogen, and subtract the enthalpy change for burning 1 mole of propane: 3(−393.5)+4(−285.8)−(−2220.0)=−1180.5−1143.2+2220.0=−103.7 kJ mol−13(-393.5) + 4(-285.8) - (-2220.0) = -1180.5 - 1143.2 + 2220.0 = -103.7\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Write the target equation for the formation of propane from its elements in standard states.
3C(s)+4H2(g)→C3H8(g)3\text{C}(s) + 4\text{H}_2(g) \rightarrow \text{C}_3\text{H}_8(g)
The enthalpy of formation definition requires forming 1 mole of propane from carbon and hydrogen gas.
2
Scale the given combustion reactions to match the stoichiometric coefficients of the target equation.
3C(s)+3O2(g)→3CO2(g)ΔH1∘=3×(−393.5)=−1180.5 kJ mol−13\text{C}(s) + 3\text{O}_2(g) \rightarrow 3\text{CO}_2(g) \quad \Delta H_1^\circ = 3 \times (-393.5) = -1180.5\text{ kJ mol}^{-1}
4H2(g)+2O2(g)→4H2O(l)ΔH2∘=4×(−285.8)=−1143.2 kJ mol−14\text{H}_2(g) + 2\text{O}_2(g) \rightarrow 4\text{H}_2\text{O}(l) \quad \Delta H_2^\circ = 4 \times (-285.8) = -1143.2\text{ kJ mol}^{-1}
3 moles of C and 4 moles of H2 are needed on the reactant side.
3
Reverse the propane combustion equation so propane appears on the product side, changing the sign of ΔH∘\Delta H^\circ.
3CO2(g)+4H2O(l)→C3H8(g)+5O2(g)ΔH3∘=+2220.0 kJ mol−13\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \rightarrow \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \quad \Delta H_3^\circ = +2220.0\text{ kJ mol}^{-1}
Reversing a reaction step changes the sign of its enthalpy change according to Hess's Law.
4
Sum the modified enthalpy values.
ΔHf∘=−1180.5+(−1143.2)+2220.0=−103.7 kJ mol−1\Delta H_f^\circ = -1180.5 + (-1143.2) + 2220.0 = -103.7\text{ kJ mol}^{-1}
According to Hess's Law, the total enthalpy change is independent of the pathway.

Key Concept

Hess's Law and Calculation of Enthalpy of Formation from Enthalpies of Combustion
Estimated Time:1m 30s
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Topics

Standard Enthalpy Changes and Hess's Law